Lemmas 1 and 2 of polywriteup.

macros.tex

@@ -21,7 +21,7 @@
 	\newcommand{\tri}{\triangle}
 	\newcommand{\twopathdis}{| \land}
 	\newcommand{\threepath}{\sqcap}
-	\newcommand{\oneint}{\Upilson}
+	\newcommand{\oneint}{\Upsilon}
 %David's Scheme
 \newcommand{\vecform}[1]{\textbf{#1}}%auxiliary cmd to use only with macros, so that we can change the format easily if need be
 \newcommand{\sone}{\vecform{x}}

poly-form.tex

@@ -94,12 +94,12 @@ The corollary follows immediately by \cref{prop:l1-rpoly-numTup}.
 \subsection{When $\poly$ is not in sum of monomials form}
 We would like to argue that in the general case there is no computation of expectation in linear time.
 
-To this end, consider the follow graph $G(V, E)$, where $|E| = m$, $|V| = \numTup$, and $i, j \in [\numTup]$.  Consider the query $q_E(\wElem_1,\ldots, \wElem_\numTup) = \sum\limits_{(i, j) \in E} \wElem_i \cdot \wElem_j$.
+To this end, consider the following graph $G(V, E)$, where $|E| = m$, $|V| = \numTup$, and $i, j \in [\numTup]$.  Consider the query $q_E(\wElem_1,\ldots, \wElem_\numTup) = \sum\limits_{(i, j) \in E} \wElem_i \cdot \wElem_j$.
 
-\begin{Lemma}
+\begin{Lemma}\label{lem:gen-p}
 If we can compute $\poly(\wElem_1,\ldots, \wElem_\numTup) = q_E(\wElem_1,\ldots, \wElem_\numTup)^3$ in T(m) time for fixed $\prob$, then we can count the number of triangles in $G$ in T(m) + O(m) time.
 \end{Lemma}
-\begin{Lemma}
+\begin{Lemma}\label{lem:const-p}
 If we can compute $\poly(\wElem_1,\ldots, \wElem_\numTup) = q_E(\wElem_1,\ldots, \wElem_\numTup)^3$ in T(m) time for O(1) distinct values of $\prob$ then we can count the number of triangles (and the number of 3-paths, the number of 3-mathcings) in $G$ in O(T(m) + m) time.
 \end{Lemma}
 
@@ -116,18 +116,54 @@ First, let us do a warm-up by computing $\rpoly(\wElem_1,\dots, \wElem_\numTup)$
 		\begin{enumerate}
 			\item First note that 
 				\begin{align*}
-					\poly_2(\wElem_1,\ldots, \wElem_\numTup) &= \sum_{(i, j) \in E} (\wElem_i\wElem_j)^2 + \sum_{(i, j), (k, \ell) \in E s.t. (i, j) \neq (k, \ell)} \wElem_i\wElem_j\wElem_k\wElem_\ell\\
+					\poly_2(\wVec) &= \sum_{(i, j) \in E} (\wElem_i\wElem_j)^2 + \sum_{(i, j), (k, \ell) \in E s.t. (i, j) \neq (k, \ell)} \wElem_i\wElem_j\wElem_k\wElem_\ell\\
 					&= \sum_{(i, j) \in E} (\wElem_i\wElem_j)^2 + \sum_{\substack{(i, j), (j, \ell) \in E\\s.t. i \neq \ell}}\wElem_i
 					\wElem_j^2\wElem_\ell + \sum_{\substack{(i, j), (k, \ell) \in E\\s.t. i \neq j \neq k \neq \ell}} \wElem_i\wElem_j\wElem_k\wElem_\ell\\
 				\end{align*}
 				By definition, 
 				\begin{equation*}
-					\rpoly_2(\wElem_1,\ldots, \wElem_\numTup) = \sum_{(i, j) \in E} \wElem_i\wElem_j + \sum_{\substack{(i, j), (j, \ell) \in E\\s.t. i \neq \ell}}\wElem_i\wElem_j\wElem_\ell + \sum_{\substack{(i, j), (k, \ell) \in E\\s.t. i \neq j \neq k \neq \ell}} \wElem_i\wElem_j\wElem_k\wElem_\ell
+					\rpoly_2(\wVec) = \sum_{(i, j) \in E} \wElem_i\wElem_j + \sum_{\substack{(i, j), (j, \ell) \in E\\s.t. i \neq \ell}}\wElem_i\wElem_j\wElem_\ell + \sum_{\substack{(i, j), (k, \ell) \in E\\s.t. i \neq j \neq k \neq \ell}} \wElem_i\wElem_j\wElem_k\wElem_\ell\label{eq:part-1}
 				\end{equation*}
 				Notice that the first term is $\numocc{\ed}\cdot \prob^2$, the second $\numocc{\twopath}\cdot \prob^3$, and the third $\numocc{\twodis}\cdot \prob^4.$
-			\item lal la\ldots
+			\item Note that 
+				\begin{align*}
+					&\numocc{\ed} = m,\\
+					&\numocc{\twopath} = \sum_{u \in V} \binom{d_u}{2} \text{where $d_u$ is the degree of vertex $u$}\\									&\numocc{\twodis} = \text{a correct formula}
+				\end{align*}
 		\end{enumerate}
+		Thus, since each of the summations can be computed in O(m) time, this implies that by \cref{eq:part-1} $\rpoly(\prob,\ldots, \prob)$ can be computed in O(m) time.\qed
 	\end{proof}
 \end{Claim}
+
+We are now ready to state the claim we need to prove \cref{lem:gen-p} and \cref{lem:const-p}.
+
+Let $\poly(\wVec) = q_E^3(\wVec)^3$.
+\begin{Claim}
+	\begin{enumerate}
+		\item\label{claim:four-one} $\rpoly(\prob,\ldots, \prob) = \numocc{\ed}\prob^2 + 6\numocc{\twopath}\prob^3 + 6\numocc{\twodis} + 6\numocc{\tri}\prob^3 + 6\numocc{\oneint}\prob^4 + 6\numocc{\threepath}\prob^4 + 6\numocc{\twopathdis}\prob^5 + 6\numocc{\threedis}\prob^6.$
+		\item\label{claim:four-two} The following can be computed in O(m) time.  $\numocc{\ed}$, $\numocc{\twopath}$, $\numocc{\twodis}$, $\numocc{\oneint}$, $\numocc{\twopathdis}$, $\numocc{\threedis}$.
+	\end{enumerate}
+	$\implies$ If one can compute $\rpoly_3(\prob,\ldots, \prob)$ in time T(m), then we can compute the following in O(T(m) + m):
+\[\numocc{\tri} + \numocc{\threepath} \cdot \prob - \numocc{\threedis}\cdot(\prob^2 - \prob^3).\]
+
+	\begin{proof}
+		By definition we have that
+		\[\poly_3(\wElem_1,\ldots, \wElem_\numTup) = \sum_{\substack{(i_1, j_1),\\ (i_2, j_2),\\ (i_3, j_3) \in E}} \prod_{\ell = 1}^{3}\wElem_{i_\ell}\wElem_{j_\ell}.\]
+		Rather than list all the expressions in full detail, let us make some observations regarding the sum.  Let $e_1 = (i_1, j_1), e_2 = (i_2, j_2), e_3 = (i_3, j_3)$.  Notice that each expression in the sum consists of a triple $(e_1, e_2, e_3)$.  There are three forms the triple $(e_1, e_2, e_3)$ can take.
+
+\underline{case 1:} $e_1 = e_2 = e_3$, where all edges are the same.  There are exactly $m$ such triples, each with a $\prob^2$ factor.
+
+\underline{case 2:}  This case occurs when there are two distinct edges of the three.  All 6 combinations of two distinct values consist of the same monomial in $\rpoly_3$, i.e. $(e_1, e_1, e_2)$ is the same as $(e_2, e_1, e_2)$.  This case produces the following edge patterns: $\twodis, \twopath$.
+
+\underline{case 3:} $e_1 \neq e_2 \neq e_3$, i.e., when all edges are distinct.  This case consists of the following edge patterns: $\threedis, \twopathdis, \threepath, \oneint, \tri$.
+
+It has already been shown previously that $\numocc{\ed}, \numocc{\twopath}, \numocc{\twodis}$ can be computed in O(m) time.  Here are the arguments for the rest.
+\[\numocc{\oneint} = \sum_{u \in V} \binom{d_u}{3}\]
+$\numocc{\twopathdis} + \numocc{\threedis} = $ the number of occurrences of three distinct edges with five or six vertices.  This can be counted in the following manner.  For every edge $(u, v) \in E$, throw away all neighbors of $u$ and $v$ and pick two more distinct edges.
+\[\numocc{\twopathdis} + \numocc{\threedis} = \sum_{(u, v) \in E} \binom{m - d_u - d_v - 1}{2}\]  The implication in \cref{claim:four-two} follows by the above and \cref{claim:four-one}.\qed
+	\end{proof}
+\end{Claim}
+\cref{claim:four-two} of Claim 4 implies that if we know $\rpoly_3(\prob,\ldots, \prob)$, then we can know in O(m) additional time
+\[\numocc{\tri} + \numocc{\threepath} \cdot \prob - \numocc{\threedis}\cdot(\prob^2 - \prob^3).\]  We can think of each term in the above equation as a variable, where one can solve a linear system given 3 distinct $\prob$ values, assuming independence of the three linear equation.  In the worst case, without independence, 4 distince values of $\prob$ would suffice...because Atri said so.
 \end{proof}